how many different digit number are there which have digit 1 2 3 4 5 6 7 8 such that

But its not mentioned anywhere in the question about construction of 4-digit number..also for number arrangement we do use permutation as far as i know...so explain
10 years agoHelpfull: Yes(17) No(1)

Digit 1 can appear in any of the 4 positions of a 4 digit number => 4C1

Further,as repetition is allowed,the othe three digits can have any of the remaining 7 numbers

So total no of different 4 digit numbers possible = 4C1 * 7 * 7 *7 = 4C1 * 7^3
10 years agoHelpfull: Yes(15) No(12)
insufficient data
10 years agoHelpfull: Yes(7) No(2)
it might be 4digit number
10 years agoHelpfull: Yes(3) No(0)
supposed that 4-digit number we have to form
1 can appear only at one place out of 4 = 4c1
now 3 places are remaining and repetition is allowed so 7*7*7
final answer : 4c1*7*7*7
10 years agoHelpfull: Yes(2) No(1)
Your question is incomplete...it should be
How many 4 digit numbers re there which have digits 1,2,3,4,5,6,7,8 such that 1 appears exactly once and repitition is allowed?
Ans=7*7*7*4c1(1 can be in unit,tens a,thousand and hundreds place)
10 years agoHelpfull: Yes(2) No(0)
it is not given its 4 digit no
10 years agoHelpfull: Yes(1) No(0)
VINAY SIPANI
why "Further,as repetition is allowed,the othe three digits can have any of the remaining 7 numbers"???
plz xplain
10 years agoHelpfull: Yes(0) No(0)

how many different digit number are there which have digit 1,2,3,4,5,6,7,8, such that digit 1 appears exactly once

its answer is 4c1 *7^3 how? why 4c1 is use here .?